Question

8. In the given figure point E is mid point of
median AD. If the produced part of BE
meets AC at F then prove that 3AF = AC.​

Answer 1

Answer:

hey mate

Step-by-step explanation:

AD is the median of ΔABC and E is the midpoint of AD

Through D

 draw DG || BF

In ΔADG

 E is the midpoint of AD and EF || DG

By converse of midpoint theorem we have

F is midpoint of AG and AF = FG  ..............1

Similarly, in ΔBCF 

D is the midpoint of BC and DG || BF   

G is midpoint of CF and FG = GC ..............2

From equations 1 and 2

we will get

AF = FG = GC ........3

 AF + FG + GC = AC

AF + AF + AF = AC (from eu 3)

3 AF = AC

AF = (1/3) AC

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